Dr. morgan is late for work. he enters the on-ramp to the merritt parkway at 15 m/s and
accelerates at a uniform 3.75 m/s2 up the 50 m ramp and for an additional 160 m on the
highway. he then sees a cop parked on the side of the road up ahead. he hits the brakes, causing him to slow at a rate of 2.50 m/s2, until he reaches the speed limit of 24.6 m/s (55 mph). determine the amount of time that passes between the moment dr. morgan
enters the on-ramp and until he reaches 55 mph (during his deceleration).

The amount of time is 14.44 seconds

Explanation:

* Lets explain how to solve the problem

- Dr.Morgan enters the on-ramp to the Merritt Parkway at 15 m/s

∴ His initial velocity u = 15 m/s

- He accelerates at a uniform 3.75 m/s² up the 50 m ramp and for

 an additional 160 m on the  highway

∴ The acceleration a = 3.75 m/s²

∴ The distance s = 50 + 160 = 210 m

- He hits the brakes, causing him to slow at a rate of 2.50 m/s²

 until he reaches the speed limit of 24.6 m/s

∴ The deceleration is 2.50 m/s²

∴ His final velocity v = 24 m/s

- We need to find the time of entire trip

* We have two stage in the problem:

# 1st stage:

- s = 210 m , u = 15 m/s , a = 3.75 m/s²

∵ s = ut + 1/2 at², where s is the distance, u is the initial velocity,

  a is the acceleration and t is the time

- Substitute the value of s , u and a in the equation above

∴ 210 = 15t + 1/2 (3.75)t²

∴ 210 = 15t + 1.875t²

- Subtract 210 from both sides

∴ 1.875t² + 15t - 210 = 0

- Solve the equation for t

∴ t = 7.31 seconds

# 2nd stage:

- d = 2.50 m/s² , v = 24.6 m/s

- The initial velocity in this stage is the final velocity of the

 1st stage, so lets calculate it

∵ v² = u² + 2as, where v is the final velocity, u is the initial velocity

   s is the distance and a is the acceleration

∴ v² = (15)² + 2(3.75)(210) = 1800

∴ v = √1800 = 42.43 m/s

∵ v in the 1st stage is u in the 2nd stage

∴ u = 42.43 m/s

- Now to fine the time of this stage we can use the rule

* Remember: d = -a

∴ seconds

∵ The time of the entire trip is the sum of the time of the 1st stage

  and the time of the 2nd stage

∵ t of 1st stage = 7.31 second

∵ t of 2nd stage = 7.13 seconds

∴ The total time = 7.31 + 7.14 = 14.44 seconds

∴ The amount of time is 14.44 seconds

the ima (higher than 1m) will be greater than the ima with the height of 1m.